Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.8 Junior Find the answer

In rectangle $ABCD, AB=5$ and $BC=3$ . Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2$ . Lines $AF$ and $BG$ intersect at $E$ . Find the area of $\triangle AEB$ :

Pick one

Solution

$\triangle EFG \sim \triangle EAB$ because $FG \parallel AB.$ The ratio of $\triangle EFG$ to $\triangle EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $\triangle EAB,$
$\frac{2}{5} = \frac{h-3}{h}$
$2h = 5h-15$
$3h = 15$
$h = 5$
The height is $5$ so the area of $\triangle EAB$ is $\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.